The Mathematics of Maybe: Introduction to ProbabilityClass 9 Mathematics NCERT Solutions

(i) The probability of an impossible event is 0.

(ii) The set of all possible outcomes of a random experiment is called the sample space.

(iii) The probability of an event that is certain to happen is 1.

(iv) Tossing a fair coin has a probability of $\frac{1}{2}$ for getting heads.

The number of students who like football is 15, and the relative frequency is $\frac{15}{50}$ (or 0.3).

Given: A bag with 3 candies: strawberry, lemon, and mint.

To Find: The probability of picking a strawberry candy.

Solution: The sample space of possible outcomes is S = {strawberry, lemon, mint}. Total number of possible outcomes = 3.

The favourable outcome is picking a strawberry candy. Number of favourable outcomes = 1.

Formula: Probability (P) = $\frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}}$

Final Answer: The probability of picking a strawberry candy is $\frac{1}{3}$.

Given: Shirts: Red (R), Blue (B) Pants: Jeans (J), Khakis (K), Shorts (S)

To Find: List all possible combinations of outfits in a table.

Solution: We can create a table where rows represent the shirts and columns represent the pants. Each cell in the table will be a unique outfit combination.

There are a total of $2 \times 3 = 6$ possible combinations.

Given: Total number of cases = 1000. Data from the table:

Formula: Probability = $\frac{\text{Number of favourable cases}}{\text{Total number of cases}}$

Solution: Number of favourable cases = 20. $P(\text{lasts < 4000 km}) = \frac{20}{1000} = \frac{1}{50} = 0.02$

Final Answer: The probability is 0.02.

Solution: This range covers two categories: '4001 to 9000' and '9001 to 14000'. Number of favourable cases = 210 + 325 = 535. $P(\text{lasts between 4000 and 14000 km}) = \frac{535}{1000} = 0.535$

Final Answer: The probability is 0.535.

Solution: Number of favourable cases = 445. $P(\text{lasts > 14000 km}) = \frac{445}{1000} = 0.445$

Final Answer: The probability is 0.445.

Given: The letters of the word 'PEACE' are on cards. The set of cards is {P, E, A, C, E}. Total number of cards = 5.

Solution: The favourable outcomes are drawing a 'P', an 'E', or a 'C'. The cards corresponding to these letters are {P, E, C, E}. Number of favourable outcomes = 4. $P(\text{P, E, or C}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{4}{5}$

Final Answer: The probability is $\frac{4}{5}$.

Solution: The favourable outcomes are drawing a card that is 'not an E'. The cards that are not 'E' are {P, A, C}. Number of favourable outcomes = 3. $P(\text{not an E}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{3}{5}$

Final Answer: The probability is $\frac{3}{5}$.

Given: A spinner with 8 equally likely outcomes: S = {1, 2, 3, 4, 5, 6, 7, 8}. Total number of possible outcomes = 8.

Formula: Probability (P) = $\frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}}$

Given: 4 Red balls (R) and 5 Blue balls (B). Total balls = 9. Two balls are drawn without replacement.

Solution: This corresponds to the path R1-B2 on the tree diagram. The probability is the product of the probabilities along this path. $P(R1 \text{ and } B2) = P(R1) \times P(B2|R1)$ $P(R1 \text{ and } B2) = \frac{4}{9} \times \frac{5}{8} = \frac{20}{72} = \frac{5}{18}$

Final Answer: The probability is $\frac{5}{18}$.

Solution: This corresponds to the path B1-B2 on the tree diagram. The probability is the product of the probabilities along this path. $P(B1 \text{ and } B2) = P(B1) \times P(B2|B1)$ $P(B1 \text{ and } B2) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72} = \frac{5}{18}$

Final Answer: The probability is $\frac{5}{18}$.

Given: A pair of 6-sided dice is thrown.

Given: A box with 4 balls numbered {1, 2, 3, 4}.

To Find: The sample space for the combined experiment.

Solution: The sample space consists of all possible pairs, with the first element from the coin toss and the second from the card draw.

Sample Space (S) = { (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6) }

The total number of elements in the sample space is $2 \times 6 = 12$.

Experiment: Tossing three coins and recording the number of heads.

The possible number of heads are 0, 1, 2, and 3. A valid sample space must list all possible outcomes and no impossible ones.

Final Answer: List (iv) {0, 1, 2, 3} is the correct sample space.

Given: A rectangular region with length 3 m and width 2 m. Inside the rectangle, there is a circle with a diameter of 1 m.

To Find: The probability that a randomly dropped dye lands inside the circle.

Concept: This is a geometric probability problem. The probability is the ratio of the favourable area to the total area. $P(\text{event}) = \frac{\text{Favourable Area}}{\text{Total Area}}$

Final Answer: The probability that the dye will land inside the circle is $\frac{\pi}{24}$.

Given: Total sweets in sample = 30 Number of red sweets = 10 Number of green sweets = 8 Number of yellow sweets = 7 Number of blue sweets = 5 Total sweets in the bag = 600

Formula: Experimental Probability = $\frac{\text{Number of times the event occurred}}{\text{Total Number of trials}}$

Solution: Number of green sweets in the sample = 8 Total number of sweets in the sample = 30 $P(\text{green sweet}) = \frac{\text{Number of green sweets}}{\text{Total number of sweets}} = \frac{8}{30} = \frac{4}{15}$

Final Answer: The probability that a randomly picked sweet from the sample is green is $\frac{4}{15}$.

Solution: First, calculate the probability of picking a yellow sweet from the sample. $P(\text{yellow sweet}) = \frac{\text{Number of yellow sweets}}{\text{Total number of sweets}} = \frac{7}{30}$ Now, use this probability to estimate the number of yellow sweets in the entire bag of 600 sweets. $\text{Estimated number of yellow sweets} = P(\text{yellow sweet}) \times \text{Total sweets in bag}$ $\text{Estimated number of yellow sweets} = \frac{7}{30} \times 600 = 7 \times 20 = 140$

Final Answer: Based on the sample, it is estimated that there are 140 yellow sweets in the large bag.

Given: Total students in sample = 40 Science Club = 14 Arts Club = 11 Sports Club = 9 Debate Club = 6 Total students in school = 800

Formula: Probability = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

Solution: Number of students who prefer the Arts Club = 11 Total number of students in the sample = 40 $P(\text{prefers Arts Club}) = \frac{11}{40}$

Final Answer: The probability that a randomly chosen student from the sample prefers the Arts Club is $\frac{11}{40}$.

Solution: First, calculate the probability that a student prefers the Sports Club based on the sample. $P(\text{prefers Sports Club}) = \frac{\text{Number of students who prefer Sports Club}}{\text{Total students in sample}} = \frac{9}{40}$ Now, estimate the number of students in the whole school who prefer the Sports Club. $\text{Estimated students for Sports Club} = P(\text{prefers Sports Club}) \times \text{Total students in school}$ $\text{Estimated students for Sports Club} = \frac{9}{40} \times 800 = 9 \times 20 = 180$

Final Answer: It is estimated that 180 students in the whole school are likely to prefer the Sports Club.

This question requires performing an experiment. The results will vary for each person who performs it. Below is a sample result and solution.

Experiment: A coin is tossed 20 times.

Sample Results: Let's assume the outcomes were: H, T, T, H, H, T, H, T, H, T, T, H, H, T, T, H, T, H, T, T

(i) How many times did you get heads? Counting the 'H's in the sample results: 9 times.

(ii) How many times did you get tails? Counting the 'T's in the sample results: 11 times. (Check: 9 + 11 = 20)

(iii) Calculate the experimental probability of getting heads. Formula: Experimental Probability = $\frac{\text{Number of times heads occurred}}{\text{Total number of tosses}}$ Solution: $P(\text{Heads}) = \frac{9}{20} = 0.45$

Final Answer: Based on this experiment, the experimental probability of getting heads is $\frac{9}{20}$ or 0.45.

(iv) If you toss the coin once more, what is the probability of getting tails? Solution: Each coin toss is an independent event. The previous results do not influence the next toss. The probability for the next toss is the theoretical probability. For a fair coin, there are two equally likely outcomes: Heads and Tails. Formula: Theoretical Probability = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$ Solution: $P(\text{Tails}) = \frac{1}{2} = 0.5$

Final Answer: The probability of getting tails on the next toss is $\frac{1}{2}$.

This question requires performing an experiment. The results will vary depending on the cup's shape and the way it is tossed. Below is a sample result and solution.

Experiment: A paper cup is tossed 100 times.

Sample Results: Let's assume the following frequencies were recorded:

Assigning Probabilities: We use the formula for experimental probability for each outcome. Formula: Experimental Probability = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

Final Answer: Based on this hypothetical experiment, the assigned probabilities are:

Given: A fair 6-sided die is rolled.

To Find: The probability of getting an even number.

Formula: Theoretical Probability (P) = $\frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}}$

Solution: The sample space (all possible outcomes) when rolling a die is S = {1, 2, 3, 4, 5, 6}. Total number of possible outcomes = 6.

The favourable outcomes (getting an even number) are E = {2, 4, 6}. Number of favourable outcomes = 3.

Now, we calculate the probability: $P(\text{even number}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{3}{6} = \frac{1}{2}$

Final Answer: The probability of getting an even number is $\frac{1}{2}$ or 0.5.

Given: Total rolls of a die = 12 Number of times '3' occurred = 3

Formula: Experimental Probability = $\frac{\text{Number of times the event occurred}}{\text{Total Number of trials}}$

Solution: $P(\text{rolling a '3'}) = \frac{3}{12} = \frac{1}{4} = 0.25$

Final Answer: The experimental probability of rolling a '3' is $\frac{1}{4}$ or 25%.

Formula: Theoretical Probability = $\frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}}$

Solution: For a fair 6-sided die, the sample space is {1, 2, 3, 4, 5, 6}. Total outcomes = 6. The favourable outcome is rolling a '3'. Number of favourable outcomes = 1. $P(\text{rolling a '3'}) = \frac{1}{6} \approx 0.167$

Final Answer: The theoretical probability of rolling a '3' is $\frac{1}{6}$ or approximately 16.7%.

Reason for difference: The experimental probability (rac{1}{4}) and theoretical probability (rac{1}{6}) are different because experimental probability is based on the results of a limited number of trials. Random chance can cause the actual outcomes to deviate from the theoretical expectation in the short term. A small number of trials (like 12) is often not enough to reflect the true underlying probability.

Expectation for more rolls: According to the Law of Large Numbers, as the number of trials increases, the experimental probability tends to get closer to the theoretical probability. Therefore, if you roll the die 60, 600, or 6000 times, you would expect the experimental probability of rolling a '3' to get progressively closer to the theoretical probability of $\frac{1}{6}$.

Given: A single 6-sided die is rolled.

To Find: The total number of possible outcomes in the sample space.

Solution: A standard 6-sided die has faces numbered from 1 to 6. When the die is rolled, any one of these numbers can be the outcome. The sample space, which is the set of all possible outcomes, is: S = {1, 2, 3, 4, 5, 6}

The total number of outcomes is the number of elements in the sample space. Total number of outcomes = 6.

Final Answer: The total number of possible outcomes in the sample space is 6.

Solution: The outcomes for rolling a die are {1, 2, 3, 4, 5, 6}. The outcomes for tossing a coin are {Heads (H), Tails (T)}. The sample space consists of all possible pairs of outcomes. S = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}

Solution: The integers 'between' -5 and +5 are all integers greater than -5 and less than +5. These integers are -4, -3, -2, -1, 0, 1, 2, 3, and 4. The sample space is: S = {-4, -3, -2, -1, 0, 1, 2, 3, 4}

Solution: The experiment is about the color of the ball drawn. The possible colors are green and red. The sample space lists the possible outcomes, which are the colors. S = {Green, Red}

Given: Snacks: Samosa (S), Pakora (P), Bhaji (B) Drinks: Chai (C), Lassi (L)

Solution: The sample space consists of all pairs where the first element is a snack and the second is a drink. S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)} Using abbreviations: S = {(S, C), (S, L), (P, C), (P, L), (B, C), (B, L)}

Solution: An event is a subset of the sample space. We need to list all outcomes from the sample space where the chosen snack is Samosa. Event E = {(Samosa, Chai), (Samosa, Lassi)} Using abbreviations: E = {(S, C), (S, L)}

Given: Basket A: 1 Apple (A), 2 Oranges (O1, O2) Basket B: 1 Banana (B), 1 Mango (M)

Since diagrams cannot be drawn, here is a text description of the tree diagram:

Solution: The sample space consists of all the possible pairs of fruits that can be picked. Let's list the outcomes from the tree diagram description. S = {(Apple, Banana), (Apple, Mango), (Orange1, Banana), (Orange1, Mango), (Orange2, Banana), (Orange2, Mango)} Since the two oranges are of the same type, we can also represent the sample space by the type of fruit: S = {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)} However, for calculating probability, it is crucial to consider each individual fruit as a distinct outcome to ensure they are equally likely. Thus, the first sample space is more accurate for calculation. Total number of outcomes = (Number of fruits in A) $\times$ (Number of fruits in B) = 3 $\times$ 2 = 6.

Formula: Probability (P) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

Solution: From the sample space S = {(A, B), (A, M), (O1, B), (O1, M), (O2, B), (O2, M)}, the total number of possible outcomes is 6. The favourable outcome is 'picking one apple and one banana', which is the pair (Apple, Banana). Number of favourable outcomes = 1. $P(\text{Apple and Banana}) = \frac{1}{6}$

Final Answer: The probability of picking one apple and one banana is $\frac{1}{6}$.

Given: 3 Red pens (R), 4 Black pens (B), 2 Green pens (G). Total pens = 3 + 4 + 2 = 9. The experiment involves two picks, with replacement.

Possible Outcomes: The possible colours for each pick are Red, Black, and Green. Since there are two picks, the outcomes are pairs of colours. Possible outcomes = {(R, R), (R, B), (R, G), (B, R), (B, B), (B, G), (G, R), (G, B), (G, G)}

Solution: First, we find the probabilities for picking each colour in a single draw. Total pens = 9. $P(R) = \frac{3}{9} = \frac{1}{3}$ $P(B) = \frac{4}{9}$ $P(G) = \frac{2}{9}$

Since the picks are independent (with replacement), we can multiply the probabilities for each sequence. The event 'both pick the same colour' consists of three mutually exclusive outcomes: (R, R), (B, B), and (G, G).

The total probability of picking the same colour is the sum of these probabilities. $P(\text{same colour}) = P(R, R) + P(B, B) + P(G, G)$ $P(\text{same colour}) = \frac{1}{9} + \frac{16}{81} + \frac{4}{81}$ To add these, we find a common denominator, which is 81. $P(\text{same colour}) = \frac{9}{81} + \frac{16}{81} + \frac{4}{81} = \frac{9 + 16 + 4}{81} = \frac{29}{81}$

Final Answer: The probability that both you and your friend pick pens of the same colour is $\frac{29}{81}$.